Evaluate the infinite geometric series: $$\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\dots$$
Answer: The series has first term $\frac{1}{3}$ and common ratio $\frac{1}{2}$, so the formula yields: $\cfrac{\frac{1}{3}}{1-\left(\frac{1}{2}\right)}=\boxed{\frac{2}{3}}$.